Let $(S, \delta )$ be as in Situation 42.7.1. Let $X$ be locally of finite type over $S$. Assume $X$ is integral and $\dim _\delta (X) = n$. Let $\mathcal{L}$ and $\mathcal{N}$ be invertible sheaves on $X$. Let $s$ be a nonzero meromorphic section of $\mathcal{L}$ and let $t$ be a nonzero meromorphic section of $\mathcal{N}$. Let $Z_ i \subset X$, $i \in I$ be a locally finite set of irreducible closed subsets of codimension $1$ with the following property: If $Z \not\in \{ Z_ i\} $ with generic point $\xi $, then $s$ is a generator for $\mathcal{L}_\xi $ and $t$ is a generator for $\mathcal{N}_\xi $. Such a set exists by Divisors, Lemma 31.27.2. Then
and similarly
Unwinding the definitions more, we pick for each $i$ generators $s_ i \in \mathcal{L}_{\xi _ i}$ and $t_ i \in \mathcal{N}_{\xi _ i}$ where $\xi _ i$ is the generic point of $Z_ i$. Then we can write
Set $B_ i = \mathcal{O}_{X, \xi _ i}$. Then by definition
Since $t_ i$ is a generator of $\mathcal{N}_{\xi _ i}$ we see that its image in the fibre $\mathcal{N}_{\xi _ i} \otimes \kappa (\xi _ i)$ is a nonzero meromorphic section of $\mathcal{N}|_{Z_ i}$. We will denote this image $t_ i|_{Z_ i}$. From our definitions it follows that
and similarly
in $\mathop{\mathrm{CH}}\nolimits _{n - 2}(X)$. We are going to find a rational equivalence between these two cycles. To do this we consider the tame symbol
see Section 42.5.
Lemma 42.27.1 (Key formula). In the situation above the cycle is equal to the cycle
\[ \sum (Z_ i \to X)_*\left( \text{ord}_{B_ i}(f_ i) \text{div}_{\mathcal{N}|_{Z_ i}}(t_ i|_{Z_ i}) - \text{ord}_{B_ i}(g_ i) \text{div}_{\mathcal{L}|_{Z_ i}}(s_ i|_{Z_ i}) \right) \]
\[ \sum (Z_ i \to X)_*\text{div}(\partial _{B_ i}(f_ i, g_ i)) \]
Proof. First, let us examine what happens if we replace $s_ i$ by $us_ i$ for some unit $u$ in $B_ i$. Then $f_ i$ gets replaced by $u^{-1} f_ i$. Thus the first part of the first expression of the lemma is unchanged and in the second part we add
(where $u|_{Z_ i}$ is the image of $u$ in the residue field) by Divisors, Lemma 31.27.3 and in the second expression we add
by bi-linearity of the tame symbol. These terms agree by property (6) of the tame symbol.
Let $Z \subset X$ be an irreducible closed with $\dim _\delta (Z) = n - 2$. To show that the coefficients of $Z$ of the two cycles of the lemma is the same, we may do a replacement $s_ i \mapsto us_ i$ as in the previous paragraph. In exactly the same way one shows that we may do a replacement $t_ i \mapsto vt_ i$ for some unit $v$ of $B_ i$.
Since we are proving the equality of cycles we may argue one coefficient at a time. Thus we choose an irreducible closed $Z \subset X$ with $\dim _\delta (Z) = n - 2$ and compare coefficients. Let $\xi \in Z$ be the generic point and set $A = \mathcal{O}_{X, \xi }$. This is a Noetherian local domain of dimension $2$. Choose generators $\sigma $ and $\tau $ for $\mathcal{L}_\xi $ and $\mathcal{N}_\xi $. After shrinking $X$, we may and do assume $\sigma $ and $\tau $ define trivializations of the invertible sheaves $\mathcal{L}$ and $\mathcal{N}$ over all of $X$. Because $Z_ i$ is locally finite after shrinking $X$ we may assume $Z \subset Z_ i$ for all $i \in I$ and that $I$ is finite. Then $\xi _ i$ corresponds to a prime $\mathfrak q_ i \subset A$ of height $1$. We may write $s_ i = a_ i \sigma $ and $t_ i = b_ i \tau $ for some $a_ i$ and $b_ i$ units in $A_{\mathfrak q_ i}$. By the remarks above, it suffices to prove the lemma when $a_ i = b_ i = 1$ for all $i$.
Assume $a_ i = b_ i = 1$ for all $i$. Then the first expression of the lemma is zero, because we choose $\sigma $ and $\tau $ to be trivializing sections. Write $s = f\sigma $ and $t = g \tau $ with $f$ and $g$ in the fraction field of $A$. By the previous paragraph we have reduced to the case $f_ i = f$ and $g_ i = g$ for all $i$. Moreover, for a height $1$ prime $\mathfrak q$ of $A$ which is not in $\{ \mathfrak q_ i\} $ we have that both $f$ and $g$ are units in $A_\mathfrak q$ (by our choice of the family $\{ Z_ i\} $ in the discussion preceding the lemma). Thus the coefficient of $Z$ in the second expression of the lemma is
which is zero by the key Lemma 42.6.3. $\square$
Remark 42.27.2. Let $(S, \delta )$ be as in Situation 42.7.1. Let $X$ be locally of finite type over $S$. Let $k \in \mathbf{Z}$. We claim that there is a complex Here we use notation and conventions introduced in Remark 42.19.2 and in addition
$K_2^ M(\kappa (x))$ is the degree $2$ part of the Milnor K-theory of the residue field $\kappa (x)$ of the point $x \in X$ (see Remark 42.6.4) which is the quotient of $\kappa (x)^* \otimes _\mathbf {Z} \kappa (x)^*$ by the subgroup generated by elements of the form $\lambda \otimes (1 - \lambda )$ for $\lambda \in \kappa (x) \setminus \{ 0, 1\} $, and
the first differential $\partial $ is defined as follows: given an element $\xi = \sum _ x \alpha _ x$ in the first term we set
where $\partial _{\mathcal{O}_{W_ x, x'}} : K_2^ M(\kappa (x)) \to K_1^ M(\kappa (x))$ is the tame symbol constructed in Section 42.5.
\[ \bigoplus \nolimits _{\delta (x) = k + 2}' K_2^ M(\kappa (x)) \xrightarrow {\partial } \bigoplus \nolimits _{\delta (x) = k + 1}' K_1^ M(\kappa (x)) \xrightarrow {\partial } \bigoplus \nolimits _{\delta (x) = k}' K_0^ M(\kappa (x)) \]
\[ \partial (\xi ) = \sum \nolimits _{x \leadsto x',\ \delta (x') = k + 1} \partial _{\mathcal{O}_{W_ x, x'}}(\alpha _ x) \]
We claim that we get a complex, i.e., that $\partial \circ \partial = 0$. To see this it suffices to take an element $\xi $ as above and a point $x'' \in X$ with $\delta (x'') = k$ and check that the coefficient of $x''$ in the element $\partial (\partial (\xi ))$ is zero. Because $\xi = \sum \alpha _ x$ is a locally finite sum, we may in fact assume by additivity that $\xi = \alpha _ x$ for some $x \in X$ with $\delta (x) = k + 2$ and $\alpha _ x \in K_2^ M(\kappa (x))$. By linearity again we may assume that $\alpha _ x = f \otimes g$ for some $f, g \in \kappa (x)^*$. Denote $W \subset X$ the integral closed subscheme with generic point $x$. If $x'' \not\in W$, then it is immediately clear that the coefficient of $x$ in $\partial (\partial (\xi ))$ is zero. If $x'' \in W$, then we see that the coefficient of $x''$ in $\partial (\partial (x))$ is equal to
The key algebraic Lemma 42.6.3 says exactly that this is zero.
Remark 42.27.3. Let $(S, \delta )$ be as in Situation 42.7.1. Let $X$ be locally of finite type over $S$. Let $k \in \mathbf{Z}$. The complex in Remark 42.27.2 and the presentation of $\mathop{\mathrm{CH}}\nolimits _ k(X)$ in Remark 42.19.2 suggests that we can define a first higher Chow group We use the supscript ${}^ M$ to distinguish our notation from the higher chow groups defined in the literature, e.g., in the papers by Spencer Bloch ([Bloch] and [Bloch-moving]). Let $U \subset X$ be open with complement $Y \subset X$ (viewed as reduced closed subscheme). Then we find a split short exact sequence for $i = 2, 1, 0$ compatible with the boundary maps in the complexes of Remark 42.27.2. Applying the snake lemma (see Homology, Lemma 12.13.6) we obtain a six term exact sequence extending the canonical exact sequence of Lemma 42.19.3. With some work, one may also define flat pullback and proper pushforward for the first higher chow group $\mathop{\mathrm{CH}}\nolimits ^ M_ k(X, 1)$. We will return to this later (insert future reference here).
\[ \mathop{\mathrm{CH}}\nolimits ^ M_ k(X, 1) = H_1(\text{the complex of Remark 0GU9}) \]
\[ 0 \to \bigoplus \nolimits _{y \in Y, \delta (y) = k + i}' K_ i^ M(\kappa (y)) \to \bigoplus \nolimits _{x \in X, \delta (x) = k + i}' K_ i^ M(\kappa (x)) \to \bigoplus \nolimits _{u \in U, \delta (u) = k + i}' K_ i^ M(\kappa (u)) \to 0 \]
\[ \mathop{\mathrm{CH}}\nolimits ^ M_ k(Y, 1) \to \mathop{\mathrm{CH}}\nolimits ^ M_ k(X, 1) \to \mathop{\mathrm{CH}}\nolimits ^ M_ k(U, 1) \to \mathop{\mathrm{CH}}\nolimits _ k(Y) \to \mathop{\mathrm{CH}}\nolimits _ k(X) \to \mathop{\mathrm{CH}}\nolimits _ k(U) \to 0 \]
Your email address will not be published. Required fields are marked.
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).
All contributions are licensed under the GNU Free Documentation License.
Comments (2)
Comment #2981 by Xia on
Comment #3105 by Johan on